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Lab Report: To Transform The Bacteria E-Coli

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Lab Report: To Transform The Bacteria E-Coli
Jennifer Hauss
March 4, 2015
Bacterial Transformation Lab Report
Introduction
In this lab, the goal was to transform the bacteria e-coli to glow in the dark (or under a black light). Four plates were set up with agar in them for the bacteria to feed on and grow. Changes were then made to the bacteria. One plate was the control plate, having only the LB or agar for the bacteria and negative pGLO, which is the liquid not containing the plasmid. This is the plate that was compared with the three others in order to identify the changes. The second plate contained negative pGLO, LB, and ampicillin. This is to see whether or not the bacteria will become resistant to the ampicillin and grow, or not be resistant and have no growth. The third …show more content…
This bacteria goes through the process of gene regulation during transcription. This process turns on only a fraction of the cell’s genes. When genes are turned off, it is because a repressor binds to an operator in the cell, which stops the process of transcription. This occurs when high levels of tryptophan bind to the repressor causing it to bind again to the operator. When a gene is turned on, a metabolite binds to the repressor, which stops it from binding to the operator and allows transcription to continue, turning the gene on. In one of these plates, the gene will be either turned on or off based on what is added to the …show more content…
One question is, which organism is better suited for total genetic transformation- single celled or multi-celled? The best answer is a single-celled organism because that one cell would be able to take up a new gene. Another question is, which would be the best choice for a genetic transformation: a bacterium, earthworm, fish or mouse? The best answer for this question would be a bacterium, since they are single-celled and they reproduce rapidly. There is also a lot of math involved. Starting with the LB/Amp/Ara plate, the amount of cells is 53 found on the plate. To then find the total amount of pGLO DNA (ug), the concentration of DNA (ug/ul) needs to be multiplied by the volume of DNA (ul). In this case .08 x 510, summing up to 40.8 ug. The next step is to find the fraction of DNA used by dividing volume spread on LB/Amp plate (ul) by the total volume in the test tube (ul). This would be 100 / 510, summing up to 10/51 or .2. After, the pGLO DNA spread (ug) must be found by multiplying the total amount of DNA used in “ug” by the fraction of DNA used. So, 40.8 x 10/51, this sums to 8.16 ug. This leads to our ultimate goal of the transformation efficiency by dividing the total number of cells growing on the agar plate by the amount of DNA spread on the agar plate. This would be 53 / 8.16, summing to 6.5 as our transformation
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